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424       XV    NORMED ALGEBRAS AND SPECTRAL THEORY

putting/, =/<pAn s tne function/, is universally measurable and bounded, and
hence fn(N) is a continuous normal operator on E which leaves Hn stable and
is zero on the orthogonal supplement of Hn. It follows now from (15.12.8)
that there is a unique unbounded normal operator whose restriction to each
Hn isfn(N), and it is this operator which is denoted byf(N). It is clear that
J(jV) = (/(AT))* I furthermore, if g : C ~C is another universally measurable
function, then again we have (f+g)(N)=f(N)+g(N) and (fg)(N) =
f(N)g(N), these equalities being interpreted in the sense explained in (15.12.1)
and (15.12.5). For the proof it is enough to consider, for m^.0 and n ^ 0,
the sets Am  of complex numbers  satisfying m g |/(Q | < m -f 1 and
n  |0(01 < n + 1; if we put Pmn = (pAmJN), Hmn = Pmn(E), fmn =f<pAm,n,
9mn =0<PAm>ni *t is c'ear that E is the Hilbert sum of the Hww, which are
stable under fmn(N) and gmn(N), and that f(N) (resp. g(N)) is the unique
normal operator whose restriction to each HmM is fmn(N) (resp. gmn(N)). Our
assertions therefore follow from the corresponding assertions for each pair

\Jrnn 5 9mn)'

PROBLEMS

Let E, F be two Hilbert spaces. A not necessarily bounded operator from E to F (or
simply an unbounded operator from E to F) is a linear mapping T of a vector subspace
dom(r) of E into F. We say that Tis closed if its graph T(T)is closed in E x F.

(a)   If 7"is closed, show that Ker(r) is closed in E. Generalize (15.12.2) to the present
situation.

(b)    If dom(r) is dense in E, define the adjoint T* of T as in (15.12.3). It is an un-
bounded operator from F to E. Generalize (15.12.4).

(c)    Suppose from now on that T7 is a closed operator from E to F, whose domain
dom( T) is dense in E. Show that the closure in E of Im(J*) is the orthogonal supple-
ment of Ker(D.

(d)    Suppose that Ker(r) = {0}, in other words that Tis injective. Then G = Im(r)
is closed in F if and only if the inverse mapping T""1 : G -> is continuous (use (a)).

(e)   Without any hypothesis on KerfT1), show that the following conditions are
equivalent: (1) Im(T) is closed in F; (2) Im(T*) is closed in E. (Let G be the closure
of Im(r*) in E. To show that (1) implies (2), consider the restriction TI of T to
G n dom(T). Show that Im^)  Im(r) and use (d) to deduce that Tr1 is continuous.
Hence show that every y e G is of the form T*  z, by remarking that the mapping
*i K^T1 ' x | y) is a continuous linear form on the Hilbert space Im(T').)

(f)   Put

the infimum being taken over the complement of Ker(r) in dom(!T) (d denotes the
distance in E). Show that Im(r) is closed in F if and only if y(T) > 0. (Express y(T)
in terms of Tr1, where TI is the operator considered in (e).) Show that y(T*) ^ y(J).follows from above that Pn = (p&n(N) is an orthogonal projection in E